Continuity, preimage of an open set of $\mathbb R^2$

Question

I'm stuck trying to find the error in the following:

. $\mathbb{R}^{2*} = \Bbb{R}^2\backslash\{(0,0)\} $ is an open set of $\mathbb{R}^{2}$

. Let $f := (\rho, \theta) \in\ U = \ (0,+\infty) \times (-\pi, \pi] \ \to (\rho \cos \theta, \rho \sin \theta) \in \ \mathbb{R}^{2*}$ the canonical polar projection. This application appears to be continuous, being the result of various operations over continuous applications.

. $f^{-1}(\mathbb{R}^{2*}) = U$ which would tend to prove that $U$ is an open set of $\mathbb{R}^{2}$...

Obviously something is wrong somewhere. :)

Thanks in advance!

P.S.: I'm French, originally, and I'm not proficient at speaking math in English. Don't hesitate telling me if I'm not saying things the right way, it can only help me get better. :)

Answer 1

Bonjour Nash,

$f$ is not defined on $\mathbb R^2$ but on $U$. You’re right that $f^{-1}(\mathbb{R}^{2*}) = U$. Which is not a contradiction because $U$ is indeed an open set OF $U$. You have to consider relative topology here.